Physical properties of haloalkanes

Bond polarity sets up the picture, but does not decide the trends. Halogens are more electronegative than carbon, so C–X is a polar covalent bond with $\delta^+$ on carbon and $\delta^-$ on the halogen. Electronegativity falls in the order F > Cl > Br > I. It is tempting to assume every physical property should track this same order — but boiling point, in fact, runs the opposite way, because a different property of the halogen is doing the real work.

Polarisability, not electronegativity, drives boiling point. Haloalkane molecules have no O–H or N–H bond, so they cannot hydrogen-bond to each other; the only force holding them together as a liquid is weak van der Waals (instantaneous dipole–induced dipole) attraction. This force strengthens as the electron cloud around an atom gets larger and more easily distorted — that is, more polarisable. Going down the halogen group, atomic size and electron count increase (F less than 3 shells → I more than 5 shells), so polarisability increases even though electronegativity decreases. The result, for a fixed alkyl group R:

where: R is the same alkyl group throughout the comparison, so the only variable is the identity of the halogen atom, and the ordering follows the growth of dispersion forces with polarisability, not the fall in electronegativity.

Three more variables shift boiling point in predictable directions. Longer chains raise boiling point (more surface for contact, more electrons overall). Branching lowers boiling point at constant molecular formula, because a compact, near-spherical molecule has less surface area touching its neighbours than an unbranched chain has. Piling more halogen atoms onto one carbon also raises boiling point, since each extra halogen adds more electrons: $\ce{CH4} < \ce{CH3Cl} < \ce{CH2Cl2} < \ce{CHCl3} < \ce{CCl4}$.

Density tracks halogen mass, not bond polarity. Heavier halogens add mass with little extra volume, so density rises I > Br > Cl > F and with the number of halogens per molecule. Bromo- and iodoalkanes, plus polyhalogenated chloroalkanes such as $\ce{CHCl3}$ and $\ce{CCl4}$, are denser than water; simple monochloroalkanes are usually lighter than water.

Solubility fails for a specific reason. A haloalkane's C–X dipole cannot form a hydrogen bond with water's O–H. Dissolving it would mean breaking part of water's strong H-bonded network without getting an equally strong replacement interaction back — energetically unfavourable. So haloalkanes are practically insoluble in water but dissolve freely in non-polar organic solvents such as benzene and ether ("like dissolves like").

Dipole moment falls gently, not sharply, down the group. $\mu(\text{C–F}) > \mu(\text{C–Cl}) \gtrsim \mu(\text{C–Br}) \gtrsim \mu(\text{C–I})$, since the electronegativity difference (which sets dipole size) shrinks from F to I, though bond-length and hybridisation effects keep the trend from being perfectly smooth.

The goal here is to explain, in one connected chain of reasoning that a student can reproduce on demand, why boiling point rises down the halogen group even though electronegativity falls down the same group — the single most-tested piece of reasoning in this topic, and the one place students most often substitute the wrong trend.

Step 1. Identify what actually holds haloalkane molecules together in the liquid state. There is no O–H or N–H bond anywhere in a haloalkane, so hydrogen bonding is off the table between two haloalkane molecules. The only attractive force available is van der Waals (London dispersion) attraction, arising from instantaneous fluctuations in electron distribution inducing a matching dipole in a neighbouring molecule.

Step 2. Recognise what controls the strength of that dispersion force: the polarisability of the electron cloud, which grows as the electron cloud becomes larger and more loosely held. Larger atoms, with electrons farther from the nucleus and more of them, are more easily distorted — hence more polarisable.

Step 3. Apply this to the halogen series. Going from F to Cl to Br to I, atomic size and total electron count increase sharply (fluorine's electrons sit close to a small nucleus; iodine's outer electrons are in a much larger, more diffuse shell). So polarisability increases in the order F < Cl < Br < I, exactly opposite to the electronegativity order.

Step 4. Separate the two ideas cleanly. Electronegativity governs how polar the C–X bond is — that is a bond property, relevant to chemical reactivity such as $S_N$ mechanisms. Polarisability governs how strong the intermolecular dispersion force is — that is what fixes boiling point, a bulk physical property. Since boiling point is a purely intermolecular-force phenomenon here (no hydrogen bonding to complicate it), it is polarisability, not bond polarity, that decides the ranking, giving R–F less than R–Cl less than R–Br less than R–I.

Question. Arrange 1-chlorobutane, 1-bromobutane, and 1-iodobutane in increasing order of boiling point, and justify the order.

Step 1. Note that all three share the identical alkyl group ($\ce{C4H9}$–), so the halogen atom is the only variable across the series — any difference in boiling point must trace back to the halogen.

Step 2. Recall that boiling point here is governed by van der Waals dispersion forces, which strengthen as the attached atom becomes larger and more polarisable, not by how electronegative that atom is.

Step 3. Rank the halogens by size and polarisability: Cl < Br < I (atomic radius and electron count both increase in this order), so dispersion forces — and hence boiling point — increase in the same order.

Step 4. State the answer with the measured values: 1-chlorobutane (78 °C) less than 1-bromobutane (101 °C) less than 1-iodobutane (130 °C).

Competitive stretch

Question. 2-chloro-2-methylpropane (tert-butyl chloride, b.p. approximately 51 °C) has a lower boiling point than its straight-chain isomer 1-chlorobutane (b.p. approximately 78 °C), even though both share the formula $\ce{C4H9Cl}$. Explain, and state which factor — halogen identity or molecular shape — is responsible here.

Step 1. Since both molecules have the same molecular formula, they have the same total electron count, so the maximum possible dispersion attraction is the same in principle — halogen identity cannot explain the gap, because there is only one halogen involved (chlorine) in both.

Step 2. The variable is molecular shape: 1-chlorobutane is an unbranched chain, while tert-butyl chloride is highly branched and nearly spherical.

Step 3. A straight chain offers more surface area for neighbouring molecules to make contact along, so the cumulative van der Waals attraction across that larger contact area is stronger; a compact, branched molecule offers less surface contact, so attraction is weaker and less thermal energy is needed to vaporise it.

Final answer: branching lowers boiling point at constant molecular formula by reducing intermolecular surface contact, so tert-butyl chloride (51 °C) boils below 1-chlorobutane (78 °C).

• Assuming electronegativity order predicts boiling-point order. Electronegativity falls F > Cl > Br > I, but boiling point rises the opposite way (F < Cl < Br < I) because dispersion forces, not bond polarity, control it.
• "Polar bond means water-soluble." The C–X dipole cannot hydrogen-bond with water, so haloalkanes stay insoluble in water despite being polar molecules.
• Ignoring branching when comparing isomers. Two haloalkanes with the same molecular formula can still have very different boiling points if one is branched — always check shape, not just formula, before ranking.
• Assuming every haloalkane is denser than water. Only bromo-, iodo-, and heavily chlorinated compounds (like $\ce{CHCl3}$, $\ce{CCl4}$) typically are; simple monochloroalkanes are usually lighter than water.

Ask the tutor: Want to practise ranking a mixed set of haloalkanes by boiling point, density, and water solubility all at once?